# effect of DMSO on pH

J. Martinez-Irujo jjmirujo at unav.es
Mon Mar 20 07:11:49 EST 2000

```Wolfgang Schechinger wrote:

>
> ...Finally some theoretical brain twisting: When you have say 50mol%
> (half of all molecules) DMSO in water, what will happen to the pH?
>
> A suggestion:
>
> Kw = ([H+]x[OH-])/[H2O] is the equilibrium constant for water
>
> If you change [H2O] (as you do when you substitute half of water
> molecules by DMSO) then [H+]x[OH-] also will have to be half. Since
> both factors are equal, [H+] and [OH-] will change absolutely by
> 1/(square root of two) (makes .707something).  Since [H+] and [OH-]
> still are equal (DMSO will not add or remove any, I assume it not
> being dissociated at all), the solution still will be neutral and the
> pH will not change by -log(.707sth) as one could suspect at a first
> glance.
>
> Would you agree?
>
> (I have no idea if there is a bug in the argumentation)
>
> Wolfgang

In my opinion there is two concepts mixed in your argumentation. First: a
neutral solution has  [H+] = [OH-] but this does not mean that neutral
solutions always have pH=7. In fact, ionization of water is an endothermic
process and ionization constant increases with temperature. Neutral water
solutions at 0, 24 and 50 ºC have pH equal to 7.473, 7.000 and 6.631,
respectively. So a pH =7 solution is neutral at 24 degrees but acid at 0
degrees and basic at 50 degrees.  Second: dielectric constant of water is
87.74 while for DMSO is 45. I would expect that H+ and OH- to interact
more in DMSO that in water. Equilibrium dissociation constant  must
decrease, and thus a relatively concentrated DMSO solution should have a
slightly alkaline pH... (only playing).

May be Bill can tell us what actually hapens in the pHmeter....
--
Juan J. Martinez Irujo
Departamento de Bioquimica